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For a ring $R$ and for $n \geq 1$, define $ S := R[x_{1},...,x_{n-1}]$ for the polynomial ring in $n-1$ variables with coefficients in $R$.

Show that $R[x_{1},...,x_{n}]$ is isomorphic to the polynomial ring $S[x_{n}]$ in one variable $x_{n}$ with coefficients in $S$

The hint I have been given is to write down a map between two rings and prove it is an isomorphism. I'm not sure how to even begin however.

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  • $\begingroup$ You have a typo, one of your n-1 should be an n. Don't think too hard about this problem, it is essentially "obvious". Once you see why it is supposed to be true, you will see what the isomorphism should be. $\endgroup$
    – Seth
    Mar 18, 2014 at 21:24
  • $\begingroup$ Thanks, I've corrected the typo. I'm afraid I really don't seem to be getting anywhere with this question $\endgroup$
    – Mathlete
    Mar 18, 2014 at 21:26

3 Answers 3

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Here's the intuition for this problem: a polynomial in $n$ variables (something in $R[x_1,\dots,x_n]$) is exactly the same as a polynomial in one variable whose coefficients are polynomials in $n-1$ variables (something in $S[x_n]$). For example, $x_1x_2^3x_3 + 4x_1x_2 + x_1x_3^2 + x_3$ is the same as $x_1\cdot x_3^2 +(x_1x_2^3+1)\cdot x_3 + 4x_1x_2$. In any case, the isomorphism is (I hope!) so obvious that there's hardly anything to write down. Formally, you could define an isomorphism $\phi:S[x_n]\to R[x_1,\dots,x_n]$ by $$ \phi\big(p_d\,x_n^d + \cdots p_1\,x_n + p_0\big) = p_d\,x_n^d + \cdots p_1\,x_n + p_0, $$ where $p_0,p_1,\dots,p_d$ are polynomials in $n-1$ variables with coefficients in $R$. It's clear that $\phi$ is an isomorphism: both injectivity and surjectivity are obvious.

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If you have a polynomial in $n$ variables you can factor out the last variable $x_n$ from every term and collect terms with like powers of $x_n$. This way you can view a polynomial in $n$ variables as a polynomial in $x_n$ over polynomials in $n-1$ variables. If you want to be formal you can use this idea to define the isomorphism.

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  • $\begingroup$ I see, I think I understand what you're getting at. I'm not sure how I'd go about writing it as a formal proof though? $\endgroup$
    – Mathlete
    Mar 18, 2014 at 21:29
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More generally, $ R [A][B]\cong R [A\sqcup B] $ for sets $ A $ and $ B $. This follows from the universal property of the polynomial ring.

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